XXXIV. All other four sided figures, besides these, are called trape ziums. XXXV. Parallel straight lines are such as are in the same plane, and which being produced ever so far both ways, do not meet. POSTULATES. I. LET it be granted that a straight line may be drawn from any one point to any other point. II. That a terminated straight line may be produced to any length in a straight line. And that a circle may be described from any centre, at any distance from that centre. AXIOMS. I. THINGS which are equal to the same are equal to one another. II. If equals be added to equals, the wholes are equal. III. If equals be taken from equals, the remainders are equal. IV. If equals be added to unequals, the wholes are unequal. V. If equals be taken from unequals, the remainders are unequal. VI. Things which are double of the same are equal to one another. VII. Things which are halves of the same are equal to one another. VIII. Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another. Book I. " If a straight line meets two straight lines, so as to make the " two interior angles on the same side of it taken together " less than two right angles, these straight lines being con"tinually produced, shall at length meet upon that side on " which are the angles which are less than two right angles. "See the notes on prop. 29. of book I." PROPOSITION I. PROBLEM. TO describe an equilateral triangle upon a given finite straight line. Let AB be the given straight line; it is required to describe an equilateral triangle upon it. Book I. Because the point A is the centre of the circle BCD, AC is equal to AB; and because the point B is the centre of the c 15. Decircle ACE, BC is equal to BA: but it has been proved that CA finition. is equal to AB; therefore CA, CB are each of them equal to AB; but things which are equal to the same are equal to one anotherd; therefore CA is equal to CB; wherefore CA, AB, BC d 1st Axare equal to one another; and the triangle ABC is therefore iom. equilateral, and it is described upon the given straight line AB. Which was required to be done. PROP. II. PROB. FROM a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line; it is required to draw from the point A a straight line equal to BC. Book I. Because the point B is the centre of the circle CGH, BC is equale to BG; and because D is the centre of the circle GKL, e 15. Def. DL is equal to DG, and DA, DB, parts of them, are equal; f 3. Ax. therefore the remainder AL is equal to the remainder f BG: but it has been shown, that BC is equal to BG; wherefore AL and BC are each of them equal to BG; and things that are equal to the same are equal to one another; therefore the straight line AL is equal to BC. Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC. Which was to be done. PROP. III. PROB. FROM the greater of two given straight lines to cut off a part equal to the less. Let AB and C be the two given straight lines, whereof AB is the greater. It is required to cut off from AB, the greater, a part equal to C, the less. D C B a 2. 1. A b 3. Post. From the point A draw a the straight line AD equal to C; and from the centre A, and at the distance AD, describe the circle DEF; and because A is the cen F c 1. Ax. tre of the circle DEF, AE shall be equal to AD; but the straight line C is likewise equal to AD; whence AE and C are each of them equal to AD: wherefore the straight line AE is equal to C, and from AB, the greater of two straight lines, a part AE has been cut off equal to C the less. Which was to be done. PROP. IV. THEOREM. IF two triangles have two sides of the one equal to two sides of the other, each to each; and have likewise the angles contained by those sides equal to one another; they shall likewise have their bases, or third sides, equal; and the two triangles shall be equal; and their other angles shall be equal, each to each, viz. those to which the equal sides are opposite. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. D Book I. AB to DE, and AC to DF; A CE F For, if the triangle ABC be applied to DEF, so that the point A may be on D, and the straight line AB upon DE; the point B shall coincide with the point E, because AB is equal to DE; and AB coinciding with DE, AC shall coincide with DF, because the angle BAC is equal to the angle EDF; wherefore also the point C shall coincide with the point F, because the straight line AC is equal to DF: but the point B coincides with the point E; wherefore the base BC shall coincide with the base EF, because the point B coinciding with E, and C with F, if the base BC does not coincide with the base EF, two straight lines would inclose a space, which is impossible. Therefore a 10. Ax. the base BC shall coincide with the base EF, and be equal to it. Wherefore the whole triangle ABC shall coincide with the whole triangle DEF, and be equal to it; and the other angles of the one shall coincide with the remaining angles of the other, and be equal to them, viz. the angle ABC to the angle DEF, and the angle ACB to DFE. Therefore, if two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal to one another, their bases shall likewise be equal, and the triangles be equal, and their other angles to which the equal sides are opposite shall be equal, each to each. Which was to be demonstrated. PROP. V. THEOR. THE angles at the base of an isosceles triangle are equal to one another; and, if the equal sides be produced, the angles upon the other side of the base shall be equal. Let ABC be an isosceles triangle, of which the side AB is C |